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  • If  98 out of  200  individuals in a population express the recessive phenotype, what percent of the population would you predict would be heterozygotes?  <div class=

If  98 out of  200  individuals in a population express the recessive phenotype, what percent of the population would you predict would be heterozygotes?
 

Option: 1

49%
 


Option: 2

42%
 


Option: 3

90%


Option: 4

70%


Answers (1)

best_answer

 

 

As we have already studied in Hardy Weinberg Equilibrium -

 According to Hardy-Weinberg equilibrium,
(p + q) 2 = 1 or  p2 + q2 + 2pq = 1 
According to this question, 
q2 (aa) = 98 
Frequency of the allele ‘a’ would be 98/200 = 0.49
Therefore, √0.49 = 0.7 = q
Hence, 1 - q = p 
             1 - 0.7 = 0.3  Therefore, p = 0.3
Now we have both p = 0.3 and q = 0.7 
Frequency of the heterozygous genotype = 2pq
Thus,                                                           = 2 x 0.3 x0.7 
                                                                     = 0.42 
Hence, the population of the heterozygotes would be 42% and the correct option is (b).
 

 

 

Posted by

Sanket Gandhi

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